GATE Questions and Answers with Explanation

Q:

Find the principal if the interest compounded at the rate of 10% per annum for two years is Rs. 420 ?

A) Rs. 2000	B) Rs. 2100
C) Rs. 1900	D) Rs. 2150

Answer & Explanation Answer: A) Rs. 2000

Explanation:

Given,
Compound rate, R = 10% per annum
Time = 2 years
C.I = Rs. 420
Let P be the required principal.
A = (P+C.I)
Amount, A = $P {(1 + (\frac{r}{100}))}^{n}$

(P+C.I) = $P {[1 + (\frac{10}{100})]}^{2}$

(P+420) = P[11/10][11/10]

P-1.21P = -420

0.21P = 420

Hence, P = 420/0.21 = Rs. 2000

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Filed Under: Compound Interest
Exam Prep: GATE , CAT , Bank Exams , AIEEE
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7 9197

Q:

A person is asked to put in a basket one apple when ordered 'one', one guava when ordered 'two', one orange whe ordered 'three' and is asked to take out from the basket one apple and one guava both when ordered 'four'. The ordered sequence executed by the person is as follows :

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4

1. How many apples will be there in the basket at the end of the above order sequence ?
(a) 4 (b) 3 (c) 2 (d) 1

2. How many fruits will be there in the basket at the end of the above order sequence ?
(a) 13 (b) 17 (c) 9 (d) 11

Answer

1. Answer (c)

Each time we order 1, one apple is added to the basket but each time when 4 is ordered, one apple is removed from it.

Required number of apples = number of 1's - number of 4's = 6 - 4 = 2

2. Answer (d)

Each time we order 1, 2 or 3, one fruit is added to the basket but when ever we order 4, two fruits are removed from it.

Required number = Number of 1's + Number of 2's + Number of 3's - 2 x (Number of 4's)

= 6 + 6 + 7 - (2x4)
= 19 - 8
= 11.

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Subject: Number, Ranking and Time Sequence Test Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk , Analyst

13 9194

Q:

Two vessel P and Q contain milk and water in ratio 3 : 2 and 5 : 3 respectively. If 10 liters of the mixture is removed from vessels P and poured in vessel Q then ratio of milk to water in vessel Q becomes 8 : 5. Find the initial quantity of water in vessel Q.

A) 10 lit	B) 6 lit
C) 16 lit	D) 12 lit

Answer & Explanation Answer: A) 10 lit

Explanation:

Given ratio of initial mixture of milk and water in Q = 5 : 3

Let the initial quantity of mixture in vessel Q = 8x

Let quantity of Milk = 5x and

Let quantity of water = 3x

According to the question,

$\frac{5 x + 10 x \frac{3}{5}}{3 x + 10 x \frac{2}{5}} = \frac{8}{5}$

=> 25x + 30 = 24x + 32

=> x = 2

Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.

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Filed Under: Ratios and Proportions
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18 9187

Q:

In how many years, Rs. 200 will produce the same interest @ 5 % as Rs. 900 produce in 2 years @ 3 ½ % ?

A) 5.2 yrs	B) 7 yrs
C) 7.9 yrs	D) 6.3 yrs

Answer & Explanation Answer: D) 6.3 yrs

Explanation:

P = Rs. 900, R = 3 ½ % = 7/2 %, T = 2 years.

Therefore,

S.I. = PTR/100

S.I. = Rs. (900 x 7/2 x 2/100) = Rs. 63

Now, P = Rs. 200, S.I. = Rs. 63, R = 5 %

Time = ((100 x 63) / (200 x 5) ) years = 6.3 years.

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Filed Under: Simple Interest
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6 9169

Q:

Two pipes A and B can fill a tank in 15 min and 20 min respectively. Both the pipes are opened together but after 4 min, pipe A is turned off. What is the total time required to fill the tank ?

A) 15 min 20 sec.	B) 16 min 40 sec.
C) 13 min 10 sec.	D) 14 min 40 sec.

Answer & Explanation Answer: D) 14 min 40 sec.

Explanation:

Part filled in 4 minutes = 4(1/15 + 1/20) = 7/15

Remaining part = 1 - 7/15 = 8/15

Part filled by B in 1 minute = 1/20

1/20 : 8/15 :: 1 : k

k = (8/15 )x 1 x 20 = 10( 2/3) min = 10 min 40 sec.

The tank will be full in (4 min. + 10 min. 40 sec) = 14 min 40 sec.

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Filed Under: Pipes and Cistern
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7 9167

Q:

An Engine length 1000 m moving at 10 m/s. A bird is flying from engine to end with 'x' m/s and coming back at '2x' m/s. Take total time of bird travelling as 187.5 s. Find x and 2x in km/hr ?

A) 21.4 and 42.8	B) 25.2 and 50.4
C) 31.4208 and 62.8416	D) 33.12 and 66.24

Answer & Explanation Answer: C) 31.4208 and 62.8416

Explanation:

Birds speeds in mtrs/sec is 'x' and '2x' .
While flying from Engine to end, relative speed = (x+10) m/sec
from end to engine, flying speed = (2x - 10) mtr/sec
so
1000/(x+10) + 1000/(2x-10) = 187.5 secs
solving it, we get
so x = 8.728 m/sec and 2x= 17.456 m/sec

x = 31.4208 km/hr and 2x = 62.8416 km/hr.

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Filed Under: Problems on Trains
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11 9147

Q:

The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. If its present value is Rs. 8,748, its purchase price was ?

A) Rs.14,000	B) Rs.12,000
C) Rs. 12,800	D) Rs. 15,000

Answer & Explanation Answer: B) Rs.12,000

Explanation:

We have P = 8748, R = -10 and n = 3
Therefore, the purchase price of the machine
= $\frac{P}{{(1 + \frac{R}{100})}^{n}}$ = $\frac{8748}{{(1 - \frac{10}{100})}^{3}}$
= 8748 x (100/90) x (100/90) x (100/90)
= 12,000.

Therefore, the purchase price of the machine was Rs. 12,000.

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Filed Under: Percentage
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7 9140

Q:

The market price of an article was 40% more than its cost price. Rana was going to sell it at market price to a customer, but he showed Rana some defects in the article, due to which Rana gave him a discount of 28.57%. Next day he came again and showed Rana some more defects, hence he gave him another discount that was equal to 12.5% of the cost price. What was the approximate profit/loss to Rana ?

A) Loss of 12.5%	B) Profit of 12.5%
C) Loss of 24.5%	D) Profit of 22.5%

Answer & Explanation Answer: A) Loss of 12.5%

Explanation:

Let the cost price be Rs. 100
Then, market price is Rs. 140
Now, the first discount is of 28.57% Rs.140 =~ Rs. 40
Hence, its selling price = 140 - 40 = Rs. 100
Now since Rana is selling at cost price, any further discount will be equal to loss percentage.
The next dicount of 12.5% will be the loss percentage to Rana.

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Filed Under: Profit and Loss
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11 9134

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